Using sA+tB to derive equation of line.

Question

So I have two points (1,4,-1) and (2,2,7).

If I subtract the first from the latter, I get the vector $i - 2j + 8k$. Combined with the first point, I have the non-parametric equation: $ x-1 = - \frac{y-4}{2} = \frac{z+1}{8}$ as the equation of the line passing through these points.

So far so good.

I'm given that if A and B are position vectors for two points (such as the points above), then the vector that represents the line through these points is $sA + tB$ where $s+t=1$.

If I choose $s=0$, then $sA + tB = 2i + 2j + 7k$. This is neither the vector nor the line equation of the above.

How does this work? What am I missing?

Answer 1

Here is how you do it.

$s(1, 4, -1) + t (2, 2, 7)$

This gives three equations.

$x = s + 2t$

$y = 4s + 2t$

$z = -s + 7t$

Let $s = 1-t$.

$x = 1+t$

$y = 4-2t$

$z = -1 +8t$

Solve for $t$ and we get $x-1 = -\frac{y-4}{2} = \frac{z+1}{8}$, as required.

Answer 2

use $s+t=1$ to write the equation of the line ...

$$\ell(t) =(1-t)A+tB = A +(B-A)t$$

Answer 3

Let two points be $ A (1,4,-1)$ and $B (2,2,7).$

$ s+t=1$. It is straightforward. With $s=0$, then $ s A + (1-s) B $ is B.With $s=1$, then $ s A + (1-s) B $ is A.